A 10 text{ inches} long pencil AB with midpoi | vedclass

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Question

(A) Let the pencil be rotated by an angle $\alpha$ about $C$. Let the new position of the pencil be $A'B'$.In the right-angled triangle formed by the

Vedclass · Accepted Answer

$	an^{-1}\left(\frac{3}{4}ight)$

Vedclass · Answer

(A) Let the pencil be rotated by an angle $\alpha$ about $C$. Let the new position of the pencil be $A'B'$.In the right-angled triangle formed by the perpendicular from $P$ to the pencil,let the angle between $PC$ and the new pencil line be $	heta$.We are given $PC = \sqrt{5}$ and the perpendicular distance $d = 1$.In the right triangle formed by $P$,the foot of the perpendicular on the pencil,and $C$,we have $\sin 	heta = \frac{d}{PC} = \frac{1}{\sqrt{5}}$.Thus,$	an 	heta = \frac{1}{2}$.Initially,the angle $\angle PCB = \phi = 	an^{-1}(2)$,so $	an \phi = 2$.The angle of rotation $\alpha$ is the difference between the initial angle $\phi$ and the new angle $	heta$.$\alpha = \phi - 	heta$.$	an \alpha = 	an(\phi - 	heta) = \frac{	an \phi - 	an 	heta}{1 + 	an \phi 	an 	heta} = \frac{2 - 1/2}{1 + 2(1/2)} = \frac{3/2}{2} = \frac{3}{4}$.Therefore,$\alpha = 	an^{-1}\left(\frac{3}{4}ight)$.

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